The key principle of optimization is this:

A quantityIf you writeucan only take a maximum or minimum value when its differential duis zero or undefined.

Be careful, because *u* might not have a maximum or minimum value!
Assuming that *u* varies continuously
(which it must if Calclulus is to be useful at all),
then it must have a maximum and minimum value
whenever the range of possibilities is *compact*;
this means that if you pass continuously through the possibilities in any way,
then you are always approaching some limiting possibility.
(In terms of *u* = *f*(*x*),
this is the case when *f* is continuous
and its domain, the range of possible values of *x*,
is a closed and bounded interval.)
However, if the range of possibilities heads off to infinity in some way,
or if there is an edge case that's not quite possible to reach,
then you also have to take a limit to see what value *u* is approaching.
If any such limit is larger than every value that *u* actually reaches
(which includes the possibility that a limit is ∞),
then *u* has no maximum value;
if any such limit is smaller than every value that *u* actually reaches
(which includes the possibility that a limit is −∞),
then *u* has no minimum value.

So in the end, you look at these possibilities:

- when the derivative of
*u*is zero or undefined, - the extreme edge cases, and
- the limits approaching impossible limiting cases.

Here is a typical problem:
The hypotenuse of a right triangle (maybe it's a ladder leaning against a wall)
is fixed at 20 feet,
but the other two sides of the triangle could be anything.
Still, since it's a right triangle,
we know that *x*^{2} + *y*^{2} =
20^{2},
where *x* and *y* are the lengths of legs of the triangle.
Differentiating this,
2*x* d*x* + 2*y* d*y* = 0.
Now suppose that we want to maximize or minimize the area of this triangle.
Since it's a right triangle,
the area is *A* = ½*x**y*,
so d*A* =
½*y* d*x* + ½*x* d*y*.
If this is zero,
then
½*y* d*x* +
½*x* d*y* =
0,
to go along with the other equation
2*x* d*x* + 2*y* d*y* = 0.

The equations at this point will always be linear in the differentials,
so think of this is a system of linear equations
in the variables d*x* and d*y*.
There are various methods for solving systems of linear equations;
I'll use the method of addition aka elimination,
but any other method should work just as well.
So ½*y* d*x* +
½*x* d*y* =
0
becomes
2*x**y* d*x* +
2*x*^{2} d*y* =
0
(multiplying both sides by 4*x*),
while
2*x* d*x* + 2*y* d*y* = 0
becomes
2*x**y* d*x* +
2*y*^{2} d*y* =
0
(multiplying both sides by *y*).
Subtracting these equations gives
(2*x*^{2} −
2*y*^{2}) d*y* =
0,
so either d*y* = 0
or *x*^{2} = *y*^{2}.
Now, *x* and *y* can change freely as long as they're positive,
but we have limiting cases:
*x* → 0^{+} and *y* → 0^{+}.
Since *x*^{2} + *y*^{2} = 400,
we see that *x*^{2} → 400, so *x* → 20,
as *y* → 0.
Similarly, *y* → 20 as *x* → 0.
In those cases, *A* = ½*x**y* → 0.
On the other hand, if *x*^{2} = *y*^{2},
then *x* = *y*,
so *x*, *y* =
10√2,
since *x*^{2} + *y*^{2} = 400.
In that case, *A* = ½*x**y* = 100.

So the largest area is 100 square feet, and while there is no smallest area, the area can get arbitrarily small with a limit of 0.

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This web page was written in 2015 by Toby Bartels, last edited on 2015 November 17. Toby reserves no legal rights to it.

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